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2 years ago

Define First Condition of Equilibrium ??

Physics

2 answers:

Harrizon [31]2 years ago

8 0

Answer:

<u>First Condition of Equilibrium :</u>

The net force in all direction must be zero

Sum of all force = 0 or net F = 0

Explanation:

Examples :

Book lyinh on table

Picture hanging on a wall

A paratroper coming down with terminal velocity

spayn [35]2 years ago

7 0

Answer:

First Condition of Equilibrium is that it must be experiencing no acceleration or the external forces acting on the body should be zero.

You might be interested in

A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards

Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

$Y = 1.6 X - 0.032x^2$

we know that maximum height occurs is given as

$x = -\frac{b}{2a}$

$y =- \frac{1.6}{2(-0.032)} = 25$

and maximum height is

$y = 1.6\times 25 - 0.032\times 25^2$

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0

3 years ago

Read 2 more answers

A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acce

erastova [34]

force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N

8 0

2 years ago

Read 2 more answers

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$